What number makes this equation true? $428 + $
$428 +{{?}}= 993$ ${428}$ ${993}$ $+?$ Let's start by adding hundreds to ${428}$ until we get as close to ${993}$ as possible without going over ${993}$. $\begin{aligned} {428} +100}=528\\\\ {528} +100}= 628\\\\ {628} +100}= 728\\\\ {728} +100}= 828\\\\ {828} +100}= 928 \end{aligned}$ If we add $5 \text{ hundreds}}$, or $5 00}$, we reach $928$. We cannot add any more hundreds without going over ${993}$. ${428}$ ${993}$ ${928}$ $+500$ Next, let's add tens to $928$ until we get as close to ${993}$ as possible without going over ${993}$. $\begin{aligned} 928 +{10}=938\\\\ {938} +{10}= 948\\\\ {948} +{10}= 958\\\\ {958} +{10}= 968\\\\ {968} +{10}= 978\\\\ {978} +{10}= 988 \end{aligned}$ If we add ${6 \text{ tens}}$, or ${60}$, we reach $988$. We cannot add any more tens without going over ${993}$. ${428}$ ${993}$ ${928}$ ${988}$ $+500$ $+60$ Finally, how many ones should we add to $988$ to get to ${993}?$ $\begin{aligned} 988+{2} &=990\\\\ 990+{3} &=993 \end{aligned}$ We add ${5\text{ ones}}$. ${428}$ ${993}$ ${928}$ ${988}$ $+500$ $+60$ $+5$ We added $5 \text{ hundreds}}$, ${6 \text{ tens}}$, and ${5\text{ ones}}$ to ${428}$ to get to ${993}$. $5 00}+{6 0}+{5}={565}$ ${428}$ ${993}$ ${928}$ ${988}$ $+500$ $+60$ $+5$ $+565$ $428 +{565}= 993$